Correct Answer: Option D

Explanation

In order to calculate this, it suffices to calculate the solubility of Ag, so as to know the amount lost.

\(K_{sp} = [Ag^{+}][Cl^{-}]\)

Let the solubilty of Ag and Cl = d

\(2 \times 10^{-10} = d \times d= d^{2}\)

\(d = \sqrt{2\times10^{-10}}\)

= \(1.414 \times 10^{-5}moldm^{-3}\)

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